[spfa] Jzoj P5869 绿洲

Description

 

Input

Output

 

Sample Input

5 7 2
2 3
1 2
1 3
1 4
2 4
3 4
3 5
4 5

Sample Output

1 2 2 1 2

 

Data Constraint

 

Hint

题解

• 每次以绿洲为起点，跑spfa，然后取max

代码

 1 #include <cstdio>
 2 #include <iostream>
 3 #include <cstring>
 4 #include <queue>
 5 using namespace std;
 6 struct edge{int to,from;}e[100010*2];
 7 int n,m,k,cnt,d[110],dis[100010],visit[100010],head[100010],ans[100010];
 8 queue<int> Q;
 9 void insert(int x,int y)
10 {
11     e[++cnt].to=y; e[cnt].from=head[x]; head[x]=cnt;
12     e[++cnt].to=x; e[cnt].from=head[y]; head[y]=cnt;
13 }
14 void spfa(int s)
15 {
16     memset(dis,-1,sizeof(dis));
17     dis[s]=0,visit[s]=1; Q.push(s);
18     while (!Q.empty())
19     {
20         int u=Q.front(); Q.pop();
21         visit[u]=0;
22         for (int i=head[u];i;i=e[i].from)
23             if (dis[e[i].to]==-1||dis[e[i].to]>dis[u]+1)
24             {
25                 dis[e[i].to]=dis[u]+1;
26                 if (!visit[e[i].to]) visit[e[i].to]=1,Q.push(e[i].to);
27             }
28     }
29 }
30 int main()
31 {
32     freopen("oasis.in","r",stdin);
33     freopen("oasis.out","w",stdout);
34     scanf("%d%d%d",&n,&m,&k);
35     for (int i=1;i<=k;i++) scanf("%d",&d[i]);
36     for (int i=1;i<=m;i++)
37     {
38         int u,v;
39         scanf("%d%d",&u,&v);
40         insert(u,v);
41     }
42     for (int i=1;i<=k;i++)
43     {
44         spfa(d[i]);
45         for (int j=1;j<=n;j++) if (dis[j]!=-1) ans[j]=max(ans[j],dis[j]);
46     }
47     for (int i=1;i<=n;i++) printf("%d ",ans[i]);
48     return 0;
49 }